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3 years ago

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If the graph of y = |x| is translated so that the point (1, 1) is moved to (4, 1), what is the equation of the new graph?

1 answer:

Kipish [7]3 years ago

4 0

First you graph all of these answers in a calculator:

y= Ix-3I is correct, it moved to (4,1).

y=IxI-3 is incorrect because it moved to (1,-2)

y=Ix +3I is incorrect because it moved to (-2,1)

y=IxI+3 is incorrect because it moved to (1,4)

So your answer is A: y= Ix-3I

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alexgriva [62]

Answer:

p(t) = 0.19e0.10t

=>p'(t) = 0.19e0.10t (0.10*1)

=>p'(t) = 0.019e0.10t

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for 2002, t=2002-1994 =8

in 2002

average price =p(8)

=>average price = 0.19e0.10*8

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rate of increase =p'(8)

=>rate of increase = 0.019e0.10*8

=>rate of increase =0.0422853... million per year

p(8)=$ 0.42 million

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3 years ago

Write an expression for the calculation double 2 and then add 5.

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The expression is (2x2)+5

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3 years ago

Analyzing a Graph In Exercise, analyze and sketch the graph of the function. Lable any relative extrema, points of inflection, a

Anit [1.1K]

Answer:

Analyzed and Sketched.

Step-by-step explanation:

We are given $y=\frac{\ln\left(5x)}{x^2}$

To sketch the graph we need to find 2 components.

1) First derivative of y with respect to x to determine the interval where function increases and decreases.

2) Second derivative of y with respect to x to determine the interval where function is concave up and concave down.

$y'=\frac{1-2\ln\left(5x)}{x^3}=0$

$x = \sqrt e/5$ is absolute maximum

$y''=\frac{6\ln\left(5x)-5}{x^4}=0$

$x=e^{5/6}/5$ is the point concavity changes from down to up.

Here, x = 0 is vertical asymptote and y = 0 is horizontal asymptote.

The graph is given in the attachment.

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3 years ago

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Nostrana [21]

1= 15
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3 years ago

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Answer:

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Step-by-step explanation:

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